Find, using calculus, the x coordinate of the turning point of the curve y=e^(3x)*cos(4x) pi/4<x<pi/2 (Edexcel C3)

The turning point of a curve is the point at which the gradient is 0 as from there it stops rising and starts falling or vice versa. To find this we differentiate y with respect to x (dy/dx) to find the general equation of the gradient of the line. As we have differentiation by parts we assign u=e^(3x) and v=cos(4x). This means du/dx = 3e^(3x) and dv/dx = -4sin(4x). Therefore y = uv and dy/dx = du/dxv+u*dv/dx. By substituting in the values found above we get dy/dx = 3e^(3x)*cos(4x) - 4e^(3x)*sin(4x).Now we have the equation for the gradient of the curve we set dy/dx = 0 for the turning point so 0 = 3e^(3x)*cos(4x) - 4e^(3x)*sin(4x). By rearranging and dividing everything by e^(3x) as it is a common factor that is never 0 we get 3cos(4x)=4sin(4x). From here we see we can make this tan(4x)=3/4 and x = 0.16088 from calculator display. As this is outside the bounds of the curve we need to find the next solution, as we are using tan(4x) we need to add pi/4 to the answer as tan(4x) repeats every pi/4 radians so x = 0.9463 to 4 dp.